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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 34

  • question_answer
    Given three vectors \[\vec{a},\vec{b}\] and \[\vec{c}\] each two of which are non-collinear. Further if \[\left( \vec{a}+\vec{b} \right)\] is collinear with \[\vec{c},\left( \vec{b}+\vec{c} \right)\] is collinear with \[\vec{a}\]and\[|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}\]. Then the value of \[\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=\]

    A) 3                     

    B)        -3

    C) 0                       

    D)        Cannot be evaluated

    Correct Answer: B

    Solution :

    [b] \[\vec{a}+\vec{b}=\lambda \vec{c}\] and \[\vec{b}+\vec{c}=\mu \vec{a}\] \[\therefore (\lambda \vec{c}-\vec{a})+\vec{c}=\mu \vec{a}(putting\,\vec{b}=\lambda \vec{c}-\vec{a})\] \[\Rightarrow (\lambda +1)\vec{c}=(\mu +1)\vec{a}\] \[\Rightarrow \lambda =\mu =-1\] \[\Rightarrow \vec{a}+\vec{b}+\vec{c}=0\] \[\Rightarrow |\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\] \[\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=-3\]


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